1. Stoichiometry Notes (Chapter 9)

2. I. Problems Involving Compounds a. Compounds are measured in molecules (or formula units) and single elements are measured in atoms. 1 mole of CO 2 = 6.022 X 10 23 molecules 1 mole of Ag = 6.022 X 10 23 atoms A mole is 6.022 X 10 23 of any substance!!!

3. Practice Problems: 1. How many atoms of carbon are in 25.0 grams of carbon? 2. How many molecules of calcium nitrate are in 0.200 moles of calcium nitrate?

4. b. Recall that the molar mass of a compound is the mass, in grams, of one mole of that compound. To determine the molar mass of a compound, add together the molar mass of all of the elements in the compound.

5. Practice problems: 3. What is the molar mass of nitrogen dioxide?

6. c.Problems involving molar mass of compounds. 4. What is the mass in grams of 1.25 moles of sulfur trioxide? 5. When bees sting they release the compound isopentyl acetate, C 7 H 14 O 2. How many molecules of isopentyl acetate are there in 2.00 grams of C 7 H 14 O 2 ?

7. II. Stoichiometry - shows the relationship between reactants and products in a chemical reaction. Now we will use our knowledge of chemical reactions and equations to solve problems.

8. a. One new conversion we need to know is the mole ratio – it relates the amounts of moles of any two substances involved in a chemical reaction. 2Al 2 O 3  4Al + 3O 2 2 mol Al 2 O 3 2 mol Al 2 O 3 4 mol Al 3 mol O 2 4 mol Al 3 mol O 2 Writing the inverse of any of the above would make a total of 6 mole ratios.

9. Practice Problem: 7. For the following chemical equation write all possible mole ratios. 2HgO  2Hg + O 2

10. b. Three types of problems involving stoichiometric calculations (mole ratios) are: 1. Mole to mole problems (1 step problems) 2. Mole to mass problems (2 step problems) 3. Mass to mass problems (3 step problems)

11. To solve these problems first write a balanced equation for the reaction. Second set-up a dimensional analysis problem, starting with the known quantity written as a fraction. Use correct mole ratios to solve the problem.

12. Practice Problems: 8. According to the following equation how many moles of lithium hydroxide are required to react with 20.0 moles of CO 2 ? CO 2 + 2LiOH  Li 2 CO 3 + H 2 O 9. How many moles of ammonia, NH 3 are produced when 6.00 moles of hydrogen gas react with excess nitrogen gas? 3H 2 + N 2  2NH 3 10. What mass, in grams, of glucose is produced when 3.00 moles of water react with carbon dioxide? 6CO 2 + 6H 2 O  C 6 H 12 O 6 + 6O 2 11. What mass in grams of magnesium oxide is produced when 2.00 moles of magnesium react with oxygen? 12. The reaction below is run using 824 grams of NH 3 and excess oxygen, how many moles of NO are formed? How many moles of H 2 O are formed? 4NH 3 + 5O 2  4NO + 6H 2 O 13. If mercury (II) oxide decomposes, how many grams of mercury (II) oxide are needed to produce 125 grams of oxygen? 14. How many grams of SnF 2 are produced from the reaction of 30.0 grams of HF with Sn? Sn + 2HF  SnF 2 + H 2 15. What mass of aluminum is produced by the decomposition of 5000. grams of Al 2 O 3 ? 16. How many molecules of hydrogen chloride gas will be produced when 33.5 grams of chlorine combine with excess hydrogen?

13. III. Percent Yield a. Theoretical yield – maximum amount of product that can be produced from a given amount of reactant. This value can be calculated from working mass to mass problems.

14. Example: In Practice Problem 15 you determined that 2646 grams of Al could be produced from 5000. grams of Al 2 O 3. This will occur if the reaction occurs without ANY type of error!!

15. b. Actual yield – actual amount of a product obtained from a reaction. It must be experimentally obtained. It cannot be more than the theoretical yield.

16. Example: Let’s say we carry out the reaction in Practice Problem 18 in a lab and we only obtain 2100. grams of Al from 5000. grams of Al 2 O 3.

17. c. Percent yield = Actual yield X 100 Theoretical yield Percent yield = 2100. grams X 100 = 79.36% 2646 grams

18. 17. When 36.8 grams of C 6 H 6 react with an excess of Cl 2, the actual yield of C 6 H 5 Cl is 38.8 grams. What is the percent yield of C 6 H 5 Cl? C 6 H 6 + C1 2  C 6 H 5 Cl + HCl Actual Yield= Theoretical Yield=

19. 18. If 75 grams of CO react to produce 68.4 grams of CH 3 OH, what is the percent yield of CH 3 OH? CO + 3H 2  CH 3 OH

20. Stoichiometry Notes (Chapter 9) PART 2

21. IV. Limiting Reactants- the reactant that limits the amounts of the other reactants that can combine and the amount of the product that can form in a chemical reaction. The substance that is not used up completely in a reaction is called the excess reactant.

22. Let’s say you need the following to make one hamburger: 2 buns 1 piece of meat 1 piece of lettuce

23. If you have 16 buns, 16 pieces of meat, and 20 pieces of lettuce, how many hamburgers can we make? What limits how many we can make? What is in excess?

24. You should be able to recognize a limiting reactant problem because they have 2 amounts given – one for each reactant! To solve these problems first write the balanced equation for the reaction. Second determine which of the reactants the limiting reactant is. (To determine which the limiting reactant is use both amounts of reactants given and setup two dimensional analysis problems to solve for the product. The smaller answer will come from the limiting reactant and will be the correct answer.)

25. Practice Problems: 19. Silicon dioxide (quartz) is usually quite uncreative but reacts readily with hydrogen fluoride according to the following equation. SiO 2 (s) + 4HF (g)  SiF 4 (g) + 2H 2 O (l) If 2 moles of HF are exposed to 4.5 moles of SiO 2, which is the limiting reactant?

26. 20. If 20.5 grams of chlorine is reacted with 20.5 grams of sodium, which reactant is in excess? 21. Using what you learned from problem #20, how much sodium chloride would be produced in 20.5 grams of chlorine is reacted with 20.5 grams of sodium? 22. Magnesium sulfate reacts with potassium. How much magnesium would be produced if 35 grams of magnesium sulfate is exposed to 45 grams of potassium?

27. V. Calculating Empirical Formulas from Chemical (Combustion) Analysis

28. When a compound containing C and H is subject to combustion with oxygen in a special combustion apparatus all the C is converted to CO 2 and the H is converted to H 2 O. The amount of C produced can be determined by measuring the amount of CO 2 produced; the amount of H produced can be determined by the amount of H 2 O produced.

29. Example: Consider the combustion of isopropyl alcohol. The sample is known to contain only C, H and O. Combustion of 0.255 grams of it produces 0.561 grams of CO 2 and 0.306 grams of H 2 O. Find the empirical formula.

30. C x H y O z + O 2  CO 2 + H 2 O 0.255g 0.561g 0.306g

31. Carbon: 12.01g C = ?g C?=0.154g C 44.01g CO 2 0.561g CO 2 Hydrogen: 2.02g H = ?g H?=0.034g H 18.02g H 2 O 0.306g H 2 O Oxygen: 0.255g = 0.154g C + 0.034g H + ? CxHyOz CH O ?=0.067g O

32. To find empirical formula remember the rhyme: % to mass, mass to mole, divide by small, multiply till whole.

33. Carbon: ?mol C  0.154g C X 1mol C = 1 12.01g C 0.0128mol/0.0042mol = 3.05 Hydrogen: ?mol H  0.034g H X 1mol H = 1 1.01g H 0.0340mol/0.0042mol = 8.095 Oxygen: ?mol O  0.067g O X 1mol O = 1 16.00g O 0.0042mol/0.0042mol = 1 Within experimental error, the most likely empirical formula is: C 3 H 8 O 1

34. Practice Problem: 23. An elemental analysis showed that an unknown sample contained only C, H, and O. A 0.1014g sample of the compound was burned in a C-H combustion train to produce 0.1486g of CO 2 and 0.0609g of H 2 O. Determine the empirical formula for this compound. 24. A 0.1153 gram sample of a pure hydrocarbon was burned in a C-H combustion train to produce 0.3986 grams of CO 2 and 0.0578 grams of H 2 O. Determine the empirical formula for this hydrocarbon.